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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

Chabot Mathematics. §5.3 GCF Grouping. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. MTH 55. 5.2. Review §. Any QUESTIONS About §5.2 → PolyNomial Multiplication Any QUESTIONS About HomeWork §5.2 → HW-12. PolyNomial Factoring Defined.

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Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

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  1. Chabot Mathematics §5.3 GCFGrouping Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

  2. MTH 55 5.2 Review § • Any QUESTIONS About • §5.2 → PolyNomial Multiplication • Any QUESTIONS About HomeWork • §5.2 → HW-12

  3. PolyNomial Factoring Defined • To factor a polynomial is to find an equivalent expression that is a product. An equivalent expression of this type is called a factorization of the polynomial • Factoring Breaks an algebraic expression into its simplest pieces • “Simplest”  Smallest Powers

  4. Example  Factoring Monomials • Find three factorizations of 24x3. • SOLUTION a) 24x3 = (6  4)(x  x2) = 6x  4x2 b) 24x3 = (6  4)(x2  x) = 6x2  4x c) 24x3 = ((−6)(−4))x3 = (−6)(−4x3)

  5. Greatest Common Factor (GCF) • Find the prime factorization of 105 & 60 • Use Factor-Tree 60 105 2  30 5  21  2 15 3  7 3  5

  6. Example  GCF • Thus • Recognize the Factors that both numbers have in COMMON • The GREATEST Common Factor is the PRODUCT of all the COMMON Factors • In This Case the GCF:

  7. Examples  GCF • Find the GCF for Monomials: 14p4q and 35pq3 • The Prime Factorizations • 14p4q = 2  7  p  p  p  p  q • 35pq3 = 5 7  p  q  q  q • Thus the GCF = 7  p  q = 7pq

  8. Examples  GCF • Find the GCF for Three Monomials:15x2 30xy2 57x3y • The Prime Factorizations • 15x2 = 3  5  x  x • 30xy2 = 2  3  5  x  y  y • 57x3y= 3  19  x  x  x  y ID the Common Factors • Thus the GCF = 3  x = 3x

  9. Factoring When Terms Have a Common Factor • To factor a polynomial with two or more terms of the form ab + ac, we use the distributive law with the sides of the equation switched: ab + ac = a(b + c). • Multiply Factor • 4x(x2 + 3x− 4) 4x3 + 12x2− 16x • = 4xx2 + 4x3x−4x4 = 4xx2 + 4x3x−4x4 • = 4x3 + 12x2− 16x = 4x(x2 + 3x− 4)

  10. Example  Factor by Distributive • Factor: 9a− 21 • SOLUTION • The prime factorization of 9a is 33a • The prime factorization of 21 is 37 • The largest common factor is 3. • 9a − 21 = 33a −37 (UNdist the 3) = 3(3a−7) • Chk: 3(3a −7) = 3 3a −3 7 = 9a − 21 

  11. Example  Factor by Distributive • Factor: 28x6 + 32x3. • SOLUTION • The prime factorization of 28x6 is • 22 7 xxxxxx • The prime factorization of 32x3 is • 22 2  2  2 xxx • The largest common factor is 22xxx or 4x3. • 28x6 + 32x3 = (4x3 7x )+ (4x3 8) = 4x3(7x3 + 8)

  12. Factor 12x5− 21x4 + 24x3 • The prime factorization of 12x5 is 2  2  3  x  x  x  x  x • The prime factorization of 21x4 is 3  7  x  x  x  x • The prime factorization of 24x3 is 2  2  2  3  x  x  x • The largest common factor is 3  x  x  x or 3x3. • 12x5– 21x4 + 24x3 = 3x3 4x2–3x3  7x + 3x3  8 =3  x  x  x 2  2 x  x =3  x  x  x 7  x =3  x  x  x 2  2 2 =3x3(4x2– 7x + 8)

  13. Example  Distributive factoring • Factor: 9a3b4 + 18a2b3 • SOLUTION • The Prime Factorizations: • The Greatest Common Factor is 9a2b3 • Distributing OUT the GCF Produces the factorization: 9a3b4 + 18a2b3 = 9a2b3(ab + 2)

  14. Example  Distributive factoring • Factor: −4xy + 8xw− 12x • SOLUTION • The Expanded Factorizations • −4xy = −4x  y • +8xw = − 2  −4x  w • − 12x = 3  −4x • Thus the Factored expression: −4xy + 8xw− 12x = −4x(y− 2w + 3)

  15. Factoring Out a Negative GCF • When the coefficient of the term of greatest degree is negative, it is sometimes preferable to factor out the −1 that is understood along with the GCF • e.g. Factor Out the GCF for Factor out only the3. Both areCorrect Or factor out the –3

  16. PolyNomial Factoring Tips • Factor out the Greatest Common Factor (GCF), if one exists. • The GCF multiplies a polynomial with the same number of terms as the original polynomial. • Factoring can always be checked by multiplying. • Multiplication should yield the original polynomial.

  17. Factoring by GROUPING • Sometimes algebraic expressions contain a common factor with two or more terms. • Example: Factor x2(x + 2) + 3(x + 2) • SOLUTION: The binomial (x + 2) is a factor of BOTH x2(x + 2) & 3(x + 2). • Thus, (x + 2) is a common factor; so x2(x + 2) + 3(x + 2) = (x + 2)x2 + (x + 2)3 = (x + 2)(x2 + 3)

  18. Grouping Game Plan • If a polynomial can be split into groups of terms and the groups share a common factor, then the original polynomial can be factored. • This method, known as factoring by grouping, can be tried on any polynomial with four or more terms

  19. Examples  Grouping • Factor by grouping. a) 3x3 + 9x2 + x + 3 b) 9x4 + 6x− 27x3− 18 • Solution a) 3x3 + 9x2 + x + 3 = (3x3 + 9x2) + (x + 3) = 3x2(x + 3) + 1(x + 3) = (x + 3)(3x2 + 1) Don’t Forget the “1”

  20. Examples  Grouping • Factor by grouping. a) 3x3 + 9x2 + x + 3 b) 9x4 + 6x− 27x3− 18 • Solution b) 9x4 + 6x− 27x3− 18 = (9x4 + 6x) + (−27x3− 18) = 3x(3x3 + 2) + (−9)(3x3 + 2) = (3x3 + 2)(3x− 9)

  21. Example  Grouping • Factor: y5 + 5y3 + 3y2 + 15 • SOLUTION y5 + 5y3 + 3y2 + 15 = (y5 + 5y3) + (3y2 + 15) Grouping Factoring each binomial = y3 (y2 + 5) + 3(y2 + 5) Factoring out the common factor(a BiNomial) = (y2 + 5)(y3 + 3)

  22. Factor 4ab + 2ac + 8xb + 4xc • Try grouping terms which have something in common. Often, this can be done in more than one way. • For example Grp-1 Grp-2 or a’s & x’s Grouping b’s & c’s Grouping

  23. Factor 4ab + 2ac + 8xb + 4xc • Next, find the greatest common factor for the polynomial in each set of parentheses. Grouping Set-1 Grouping Set-2 • The GCF for (4ab + 2ac) is 2a • The GCF for (8xb + 4xc) is 4x • The GCF for (4ab + 8xb) is 4b • The GCF for (2ac + 4xc) is 2c

  24. Factor 4ab + 2ac + 8xb + 4xc • Write each of the polynomials in parentheses as the product of the GCF and the remaining polynomial • Apply the distributive property to any common factors

  25. Factor 4ab + 2ac + 8xb + 4xc • Examine the Factorizations • Notice that it did not matter how the terms were originally grouped, the factored forms of the polynomials are IDENTICAL

  26. WhiteBoard Work • Problems From §5.3 Exercise Set • 22, 32, 52, 56, 68, 84 • Factor byGrouping

  27. All Done for Today Factoring4-TermPolynomials

  28. Chabot Mathematics Appendix Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu –

  29. Graph y = |x| • Make T-table

  30. Factor 4ab + 2ac + 8xb + 4xc • Divide each polynomial in parentheses by the GCF

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